> [!NOTE] Definition/ Theorem (Euler-Mascheroni constant) >Euler's constant $\gamma$ is given by the [[Series of Real Sequence|series]] $\gamma = \sum_{n=1}^{\infty} \left( \frac{1}{n} - \log\left( 1+ \frac{1}{n} \right) \right) $Equivalently $\gamma = \lim_{ m \to \infty } \left( -\log(m+1) + \sum_{n=1}^{m} \frac{1}{n} \right) $ > >*Proof*. We prove that the series converges and that the definitions are equivalent. >Note that the [[Real Natural Logarithm Function#^badf18|logarithm]] satisfies the following inequalities $\log(1+x)\leq x$for all $x>-1$ and $x-\frac{x^{2}}{2}\leq \log(1+x)$for all $x\geq 0.$ Setting $x=\frac{1}{n}$ where $n\in\mathbb{N}$ gives $0 \leq \frac{1}{n} \leq \frac{1}{n} - \log\left( 1 + \frac{1}{n} \right) \leq \frac{1}{2n^{2}}$so the series converges by [[Comparison Test for Series With Non-negative Terms (Corollary 1)|comparison]] with $\sum \frac{1}{n^{2}}.$ > >We already know that the partial sums converge to $\gamma$: $\sum_{n=1}^{m} \left( \frac{1}{n} - \log\left( 1+ \frac{1}{n} \right) \right) \to \gamma \quad \text{as } m\to \infty$so STP $\log(m+1) = \sum_{n=1}^{m} \log \left( 1+ \frac{1}{n} \right)$but $\sum_{n=1}^{m} \log \left( 1+ \frac{1}{n} \right) = \sum_{n=1}^{m} \log(n+1) - \log n = \log(m+1) -\log 1$because the sum [[Telescoping Sum|telescopes]]. > [!info] Lemma (Numerical value) > The first few digits of $\gamma$ are as follows: $\gamma \approx 0.57721566490153286060651209008240243\dots$